Math 147 Section 3.4 Inverse of a Square Matrix Matrix Equations Determinants of Matrices 1 Application Example Set up the system of equations and then solve it by using an inverse matrix. One safe investment pays 10% per year, and a more risky investment pays 18% per year. A woman has $149,600 to invest and would like to have an income of $20,000 per year from her investments. How much should she invest at each rate? 2 1
Identity Matrix Recall the identity matrix, I, is a square matrix with 1s down the diagonal and 0s elsewhere. For any square matrix, A, of the same order as I, 3 Inverse Matrix Now we introduce the inverse matrix: Two square matrices A and B are inverses of each other if: Notation: B = A 1 and A = B 1 so (A 1 ) 1 = A 4 2
Inverse Matrix Example: 1 2 1 2 1 0 1 1 1 1 0 1 1 2 1 2 1 0 1 1 1 1 0 1 5 Consider the matrix Inverse Matrix A 1 1 9 1 a matrix A that satisfies A 1 A = I. 1 a b c d a b1 1 c d 9 1. The inverse of A is 1 0 0 1 Multiply out, set it equal to,and equate corresponding entries to come up with a system of 4 equations for the unknowns a, b, c, and d. Then solve that system of unknowns using basic algebra. Use the values you obtain to write down the inverse matrix:a 1. 6 3
Inverse Matrix a b1 1 a9b ab 1 0 c d 9 1 c9d cd 0 1 7 Inverse Matrix We use elementary row operations on augmented matrices to find the inverse of a square matrix: To find the inverse of the square matrix nxn A: 1 Form the augmented matrix [A I], I is the nxn identity matrix 2 Perform elementary row operations until you get an augmented matrix of the form [I B] If A has no inverse, the reduction process will give a row of 0s in the left half of the matrix 3 The matrix B is the inverse of A 8 4
Inverse Matrix 1 2 1 0 1 1 0 1 1r1 + r2 r2 1r2 r2 2r2 + r1 r1 9 4 7 1 0 1 2 0 1 1/4r1 r1 Find the Inverse Matrix 4 7 1 2 7/4r2 + r1 r1 1r1 + r2 r2 4r2 r2 10 5
Matrix Equations Consider the system of equations: 2x + 5y + 4z = 4 x + 4y + 3z = 1 x 3y -2z = 5 Two matrices are equal if they are the same order and each entry in one is equal to its corresponding entry in the other so: 2x5y4z 4 x 4y 3z 1 x3y2z 5 Each is a 3 x 1 matrix and corresponding elements are equal. 11 Matrix Equations Lets factor the left hand matrix: 2x5y4z 4 x 4y 3z 1 x3y2z 5 12 6
Matrix Equations Now let s multiply both sides of this equation by the inverse of the first matrix. 1 2 5 4 1 2 1 1 4 3 5 8 2 1 3 2 7 11 3 1 2 1 2 5 4 x 1 2 1 4 5 8 2 1 4 3 y 5 8 2 1 7 11 3 1 3 2 z 7 11 3 5 13 Matrix Equations 1 0 0 x 3 0 1 0 y 2 0 0 1 z 2 OR Inverse matrices can be used to solve systems of equations if the system has a unique solution. System AX = B has a unique solution if and only if A 1 exists. 14 7
Inverse of 2x2 Matrix a A c b d 1 1 d b A ad bc c a Provided ad bc 0. If ad bc = 0, A 1 does not exist. 15 Determinants of Matrices The determinant of a 2 x 2 matrix is a b a b DET ad bc c d c d notation is like matrix but without the hooks on the lines 16 8
Determinants of Matrices For 3 x 3 matrix a b c d e f g h i 17 Determinants of Matrices If the determinant of a matrix B is zero, then B 1 does not exist, and if B is the coefficient matrix of a system of equations, there is no unique solution to the system. 18 9
Matrix Equations Example: using the calculator. 2x y 2z = 2 3x y + z = 3 x + y z = 7 Coefficient Matrix: Determinant: 2 1 2 3 1 1 1 1 1 19 Matrix Equations (cont.) Example: using the calculator. 2x y 2z = 2 3x y + z = 3 x + y z = 7 20 10
Solve System Using Matrix Equation x + y + 2z = 8 2x + y + z = 7 2x + 2y + z = 10 21 Application Example One safe investment pays 10% per year, and a more risky investment pays 18% per year. A woman has $149,600 to invest and would like to have an income of $20,000 per year from her investments. How much should she invest at each rate? x is $ in safe investment; y is $ in riskier investment. 22 11
Example A product is made by only two competing companies. Suppose Company X retains two-thirds of its customers and loses one-third to Company Y each year, and Company Y retains three-quarters of its customers and loses one-quarter to Company X each year. We can represent the number of customers each company had last year by: x0 y 0 where x 0 is the number Company X had and y 0 is the number Company Y had. 23 Example The number that each will have this year can be represented by 2 1 x 3 4x0 y 1 3 y 0 3 4 If Company X has 1900 customers and Company Y has 1700 customers this year, how many customers did each have last year? 24 12
Example 2 1 1900 3 4x 0 1700 1 3 y 0 3 4 25 13